Show that a permutation is odd iff it has an odd number of even cycles.
Solution
Write \(\sigma\in S_n\) as a product \(\sigma=c_1c_2\cdots c_r\) of disjoint cycles of lengths \(\ell_1,\dots,\ell_r\), and let \(E=\{i:\ell_i\text{ is even}\}\).
1. A cycle of length \(\ell\ge1\) has sign \((-1)^{\ell-1}\).
For \(\ell\ge2\), \[(a_1\ a_2\ \dots\ a_\ell)=(a_1\ a_\ell)(a_1\ a_{\ell-1})\cdots(a_1\ a_2)\] is a product of \(\ell-1\) transpositions, and \(\operatorname{sgn}\) is a homomorphism sending each transposition to \(-1\). For \(\ell=1\) the cycle is the identity, of sign \(1=(-1)^0\).
2. \(\operatorname{sgn}(\sigma)=(-1)^{|E|}\).
By step 1 and multiplicativity of \(\operatorname{sgn}\), \[\operatorname{sgn}(\sigma)=\prod_{i=1}^r(-1)^{\ell_i-1}=(-1)^{\sum_i(\ell_i-1)}.\] The integer \(\ell_i-1\) is odd exactly when \(i\in E\), so \(\sum_i(\ell_i-1)\equiv|E|\pmod 2\).
3. Q.E.D.
By step 2, \(\sigma\) is odd if and only if \((-1)^{|E|}=-1\), that is, if and only if \(|E|\) is odd.